98 Normat 59:3-4, 98–116 (2011)
What you should know about cubic and
quartic equations
Juliusz BrzeziÒski
Mathematical Sciences
Chalmers and Gothenburg University
S–41296 Göteborg, Sweden
jub@chalmers.se
1 Introduction
Polynomial algebraic equations of low degrees in the context of their Galois groups
are discussed in many places. A very long time ago, the supervisor of my Master
Thesis, the great logician, professor Andrzej Mostowski urged me to write a text
concerning this topic in order to answer several questions from the students attend-
ing his class on Galois theory in which I was a teaching assistant. The idea was to
give a straightforward, simple and direct description suitable for undergraduates.
I eagerly accepted my first serious pedagogical mission. The text, far from perfect,
appeared in Polish in [B]. During the years which followed, I taught Galois theory
several times and very often was asked similar questions as during my first teaching
task. I often xeroxed a table summarizing all cases from my pap er (correcting s ome
printing errors in it), but I couldn’t give a simple and easily accessible reference
explaining the results. Now there are many sources which could be recommended,
notably [C], and several others. But in spite of the fact that there exist different
presentations of this topic, it seems that there are many people who are in one way
or another dissatisfied with the existing texts. So here is one more with the hope
that simplicity of the formulations and arguments will give a useful pedagogical
reference.
We intend to work over an arbitrary field K of characteristic 0, but one can think
of K as any number field
1
, e.g. the field of rational numbers Q. We shall study
particular polynomial equations f(x)=0with coefficients in K. The set of all such
polynomials f(x) will be denoted, as usual, by K[x].
Let f(x) œ K[x] be a polynomial of degree n and let –
1
,–
2
,...,–
n
denote the
solutions of the equation f(x)=0in a field containing the field K. If the polynomial
has number coefficients, we can take the complex numbers as such a field containing
K. However, if K is an arbitrary field, we can fix a field K containing K in which
1
The results are exactly the same when K is an arbitrary field whose characteristic is neither
2or3,butonehastobealittlemorecautiouswithformulationsassumingonlythis.
Normat 3-4/2011 Juliusz BrzeziÒski 99
every polynomial over K can be factorized into a product of linear factors, that is,
for every polynomial f(x) of degree n with coefficients in K,thereexist–
1
,...,–
n
œ
K such that f(x)=a(x ≠ –
1
) ···(x ≠ –
n
),wherea œ K. For a proof of existence
of such a field, see [L], p. 231.
The splitting field of a polynomial f (x) over K is the field K(–
1
,–
2
,...,–
n
),
which is the least set obtained from the elements of K and the solutions –
1
,–
2
,...,–
n
of f(x)=0in K by means of the four arithmetical operations (addition, subtrac-
tion, multiplication and division). Sometimes, this field will be denoted by K
f
.
Solving the polynomial equations or describing their Galois groups over a field K,
we shall often assume that the polynomials (of degree at least 2), which we discuss,
are irreducible, that is, can not be represented as a product of two nonconstant
polynomials of lower degrees with coefficients in K. In practice, it is a natural
assumption if we know how to handle the polynomials of degrees lower than the
degree of f(x), since a nonconstant reducible polynomial is a product of two poly-
nomials of lower degrees. The sim plest possibility is that f(x) has a zero – œ K.
Then, of course, f(x)=(x ≠–)f
1
(x),wheref
1
(x) œ K[x] has the degree one lower
than the degree of f(x). A polynomial is called separable if its zeros are different.
We shall always assume that irreducible polynomials have this property. This is the
case over the fields of characteristic 0 (like the number fields) and over the finite
fields.
If f(x) œ K[x] is a polynomial with coefficients in K, then its Galois group (one
may say, the Galois group of the equation f(x)=0) is the automorphism group
of the splitting field K
f
over K. This group will be denoted by G(K
f
/K).If
‡ œ G(K
f
/K),then
‡(– + —)=‡(–)+‡(—),‡(–—)=‡(–)‡(—) and ‡(a)=a
for any –, — œ K
f
and a œ K. Hence ‡(g(–)) = g(‡(–)) for every polynomial
g œ K[x] and every – œ K
f
. In particular, we have ‡(f(–
i
)) = f(‡(–
i
)) = 0 for
i =1, 2,...,n,so‡(–
i
) is also a solution of the equation f (x)=0(notice that
‡(0) = 0). Therefore, an automorphism ‡ gives a pe rmutation of the zeros –
i
of
f(x)=0. Such a permutation defines uniquely the effect of ‡ on all the elements of
K
f
. When we have fixed a numbering of the zeros of f (x), then every automorphism
‡ gives a permutation: to ‡(–
i
)=–
‡(i)
corresponds the permutation mapping i
onto ‡(i) for i =1, 2,...,n. Usually, we shall identify the permutation of the –
i
with the permutation of their indices i and we shall consider the corresponding
permutation group as the Galois group of the polynomial f (x). One of the main
results of Galois theory is a one-to-one corresp ondence between the subgroups of the
Galois group G(K
f
/K) and the subfields of K
f
containing K:IfH is a subgroup of
G(K
f
/K) then the corresponding subfield consists of all – œ K
f
for which ‡(–)=
– for each ‡ œ H. Conversely, to a subfield L of K
f
containing K corresponds the
subgroup consisting of all ‡ œ G(K
f
/K) for which ‡(–)=– for each – œ L (the
Galois group G(K
f
/L)).
As a permutation group, the Galois group G(K
f
/K) of a polynomial f(x) of degree
n is a subgroup of the symmetric group S
n
consisting of all p e rmutations of the
100 Juliusz BrzeziÒski Normat 3-4/2011
numbers 1, 2,...,n. As we know, the group S
n
has n! elements. For every particular
polynomial f(x) of degree n, we get a particular subgroup of S
n
. A description
of these subgroups for arbitrary polynomials is not easy (and in full generality,
probably, impossible), but for polynomials of law degrees, in particular, for cubic
and quartic polynomials, the task is not too difficult and its solution is the main
purpose of this text. Of course, it is necessary to identify different subgroups of S
3
and S
4
in order to identify different cases. Such a discussion of S
3
and S
4
is given
in the Appendix.
Let f(x) be a polynomial of degree n over K with splitting field K
f
. Let –
1
,–
2
,...,–
n
be all zeros of f(x) in K
f
.Thediscriminant of f(x) is defined as the product
(1) (f)=
Ÿ
1Æi<jÆn
(–
i
≠ –
j
)
2
.
Rudimentary Galois theory says that an element of the splitting field K
f
belongs
to K if and only if it is fixed by all automorphism belonging to the Galois group
G(K
f
/K). Therefore (f ) must be in K, since, evidently, it is fixed by all possible
permutations of –
i
, i =1, 2,...,n. In fact, it is always possible to express the
discriminant as a polynomial of the coeffic ients of f(x) (see below such formulae
for polynomials of degrees Æ 4 and the comments on checking their validity in the
next section – see 2.1).
After these introductory remarks, we look at the cases, which we want to discuss
– the cubic and quartic polynomials.
A general cubic polynomial can be written in the form a
3
x
3
+ a
2
x
2
+ a
1
x + a
0
,
where a
i
œ K and a
3
”=0.Dividingbya
3
, we get a polynomial with the same zeros,
so we can always assume that a
3
=1. It is also very easy to achieve a
2
=0. In fact,
if already a
3
=1, we can always choose y, such that x = y ≠
a
2
3
.Substitutingthis
x gives an equation in which the coefficient of y
2
is 0. Thus, we shall only consider
cubic equations:
f(x)=x
3
+ px + q =0,(2)
with p, q œ K. How to solve arbitrary cubic equations in general case is presented
in Section 2 (see 2.11). If the given cubic polynomial f(x) is reducible, that is, it
can be factorized as a product of a linear and a quadratic factor over K,thenit
is easy to solve the equation f(x)=0and to find the Galois group G(K
f
/K).We
discuss in Section 2 (see 2.4) how to decide whether f(x) is reducible or not. Now
we describe the Galois groups of irreducible cubic polynomials. They are subgroups
of the permutation group S
3
of all permutations of its 3 z eros and are given in the
following simple way (for a proof see 2.11 and 2.1 for the discriminant formula):
Theorem 1.1. Let f(x)=x
3
+ px + q œ K[x] be an irreducible polynomial of
degree 3 and let =(f)=≠4p
3
≠ 27q
2
. Then
Normat 3-4/2011 Juliusz BrzeziÒski 101
G(K
f
/K)=
;
S
3
when
Ô
”œ K,
A
3
when
Ô
œ K.
A general quartic equation has a shape a
4
x
4
+ a
3
x
3
+ a
2
x
2
+ a
1
x + a
0
=0,where
a
i
œ K and a
0
”=0. As for the cubics, we may divide by a
4
and assuming a
4
=1,
we achieve a
3
=0substituting this time x = y ≠
a
3
4
. This gives an equation in
which the coefficient of y
3
is 0. Thus, we shall only consider equations:
f(x)=x
4
+ px
2
+ qx + r =0,(3)
with p, q, r œ K.
How to solve a general quartic equation, we discuss in 2.12. Studying the Galois
group in this case, like in the case of cubics, first of all we would like to know
whether the equation has a solution in K. If so, then we essentially have to handle
a cubic equation when we want to solve (3) or to determine its Galois group, since
f(x) is then a product of a linear and a cubic factor over K. For a discussion how
to decide whether f(x)=0has a solution in K see 2.4.
Thus, in principle, the problem is reduced to the case when a quartic polynomial
does not have zeros in K, so it can not be factored as a product of a first degree
polynomial and a cubic polynomial. But it may be possible to factor f(x) as a
product of two quadratic polynomials. In fact, studying possibility to factorize a
quartic polynomial as a product of two quadratic polynomials is not only a key to
a solution of the equation f(x)=0, but also to the description of its Galois group.
So let us consider a factorization
(4) x
4
+ px
2
+ qx + r =(x
2
+ ax + b)(x
2
+ a
Õ
x + b
Õ
),
where a, b, a
Õ
,b
Õ
may belong to a field containing K. Compering the coefficient for
x
3
to the right and to the left, we see immediately that a + a
Õ
=0,soa
Õ
= ≠a.
Then compering the remaining coefficients, we get the system
b + b
Õ
= a
2
+ p,
a(b
Õ
≠ b)=q,(5)
bb
Õ
= r.
Multiply the first equation by a, then square the first two equations and finally
subtract the second from the first! We get
4a
2
bb
Õ
= a
2
(a
2
+ p)
2
≠ q
2
.
102 Juliusz BrzeziÒski Normat 3-4/2011
Replacing bb
Õ
from the last equation gives our final result:
a
6
+2pa
4
+(p
2
≠ 4r)a
2
≠ q
2
=0.
The polynomial
(6) r(f)(t)=t
3
+2pt
2
+(p
2
≠ 4r)t ≠ q
2
is called the resolvent of f (x). One of its zeros is t = a
2
. In general, the resolvent
is responsible for the possibility to split f(x) into a product of two quadratic
polynomials. We can formulate this as follows in a special case when we urge that
the factorization is already over K:
Proposition 1.1. Let f(x)=x
4
+px
2
+qx+ r be a polynomial with coefficients in
a field K without zeros in K. Then f is reducible in K (a product of two quadratic
polynomials with coefficients in K) if and only if
(a) q ”=0and the resolvent r (f ) has a zero, which is a square in K,
or
(b) q =0and the resolvent r(f) has two zeros, which are squares in K or ” = p
2
≠4r
is a square in K.
For a proof see 2.5.
Remark 1.1. Notice that the case (b) is concerned with the biquadratic polyno-
mials. Notice also that q =0if and only if the resolvent r(f)(t)=t
3
+2pt
2
+(p
2
≠
4r)t ≠ q
2
of f(x)=x
4
+ px
2
+ qx + r =0has a zero t =0.
Once we know that the polynomial f(x) is reducible (but without zeros in K), we
can easily describe its Galois group (see 2.9):
Theorem 1.2. Let f(x)=x
4
+ px
2
+ qx + r be reducible in K but without zeros
in K. Then
G(K
F
/K)=
;
V
4
when r(f) has only one zero in K,
C
2
when r(f) has all its zeros in K.
The Galois group of an irreducible quartic is given in the following way (for a proof
see 2.8 and 2.1 for the discriminant formula):
Theorem 1. 3. Let K be a field and f(x)=x
4
+px
2
+qx+ r œ K[x] an irreducible
polynomial. Let
r(f)(t)=t
3
+2pt
2
+(p
2
≠ 4r)t ≠ q
2
Normat 3-4/2011 Juliusz BrzeziÒski 103
be the resolvent of f and
=(f)=(r(f)) = ≠4p
3
q
2
≠ 27q
4
+ 16p
4
r ≠ 128p
2
r
2
+ 144pq
2
r + 256r
3
its discriminant. Let ” = p
2
≠ 4r. Then:
G(K
F
/K)=
Y
_
_
_
_
_
_
_
_
_
]
_
_
_
_
_
_
_
_
_
[
S
4
when r(f) does not have zeros in K and
Ô
”œ K,
A
4
when r(f) does not have zeros in K and
Ô
œ K,
D
4
when r(f) has only one zero — œ K and
— ”œ K
if — ”=0or
Ô
” ”œ K if — =0,
C
4
when r(f) has only one zero — œ K and
— œ K
if — ”=0or
Ô
” œ K if — =0,
V
4
when r(f) has all its zeros in K.
Example. Let us find the Galois group of the polynomial f(x)=x
4
+4x ≠ 1
over the rational numbers. Is the polynomial f(x) reducible or irreducible over Q?
First we test possibility that f(x) has a rational zero. According to 2.4, we look
at the divisors of 1 (that is, ±1) and check that f(±1) ”=0.Thusf(x) can not be
factored as a product of rational polynomials of degrees 1 and 3. Then we want to
test possibility that f(x) is a product of two quadratic polynomials. According to
2.5, we look at the resolvent r(f)(x)=x
3
+4x ≠ 16. Testing the divisors of 16,
we find that — =2is the only rational ze ro of the resolvent and ≠1 ± i
Ô
7 are the
remaining two. Thus, we know that the polynomial f(x) is irreducible according
to 2.5, since
Ô
— ”œ Q. Now, we can use Theorem 1.3, which says that the Galois
group dep e nds on the product —,where=≠7168. Of c ourse,
— ”œ Q,so
the Galois group of f(x) is D
4
. 2
2 Proofs
2.1 How to compute the discriminant?
The discriminant of a polynomial f(x) with coefficients in K whose all zeros are
–
1
,...,–
n
(in some field K containing K) was defined as
(7) (f)=
Ÿ
1Æi<jÆn
(–
i
≠ –
j
)
2
.
If f(x)=x
2
+ px + q,then
(f)=(–
1
≠ –
2
)
2
=(–
1
+ –
2
)
2
≠ 4–
1
–
2
= p
2
≠ 4q,
104 Juliusz BrzeziÒski Normat 3-4/2011
since –
1
+ –
2
== ≠p and –
1
–
2
= q.
If f(x)=x
3
+px+q, then similar, but much more complicated, computations show
that
(f)=(–
1
≠ –
2
)
2
(–
2
≠ –
3
)
2
(–
3
≠ –
1
)
2
= ≠4p
3
≠ 27q
2
taking into account the Vietâ formulae –
1
+ –
2
+ –
3
=0, –
1
–
2
+ –
2
–
3
+ –
3
–
1
= p,
–
1
–
2
–
3
= ≠q.
Still more complicated is a computation showing that for f(x)=x
4
+ px
2
+ qx+ r,
(f)=
Ÿ
1Æi<jÆ4
(–
i
≠ –
j
)
2
= ≠4p
3
q
2
≠ 27q
4
+ 16p
4
r ≠ 128p
2
r
2
+ 144pq
2
r + 256r
3
.
A detailed computation in the case of cubic equations can be found in [N], p. 188.
More general considerations and formulae, which explain very well the notion of
discriminant and a method, which gives a possibility to express it by means of the
coefficients of the equation, can be found in [T], pp. 167–170. Today, it is easy
to use different symbolic computer packages like Maple, Pari or Mathematica in
order to get these formulae and compute discriminants of many other interesting
polynomials. However, those given above is all what we need in order to find Galois
groups in particular cases and we only need the definition of the discriminant
together with its fundamental property in Proposition 2.1 in order to prove all
results on Galois groups in this text.
2.2 What is the role of the discriminant?
First of all, the discriminant of a polynomial tells us whether its zeros are different.
This follows from the very definition (7), which easily implies that (f) ”=0if and
only if all the zeros of f (x) are different. The following property of the discriminant
is very important in applications related the the study of the Galois groups of
polynomials:
Proposition 2.1. Let =(f) be the discriminant of the polynomial f(x). Then
Ô
œ K if and only if all permutations in G(K
f
/K) are even.
Proof. By the definition (7), we have
Ô
œ K
f
. If there is an odd permutation in
the Galois group G(K
f
/K), then it acts on
Ô
œ K
f
by the change of its sign, so
Ô
can not be in K. If all permutations in the Galois group G(K
f
/K) are even,
then
Ô
œ K
f
is mapped on itself by all off them. Hence, we have
Ô
œ K.
2.3 The resolvent of a quartic – some properties
Lemma 2.1. If –
1
,–
2
,–
3
,–
4
are all the zeros of a quartic polynomial f(x)(in a
field K ´ K), then —
1
=(–
1
+ –
4
)
2
,—
2
=(–
2
+ –
4
)
2
,—
3
=(–
3
+ –
4
)
2
are all the
zeros of its resolvent.
Normat 3-4/2011 Juliusz BrzeziÒski 105
Proof. As we noted before, depending on the choice of the factorization (4), the
zeros of the first factor can be taken as –
1
,–
4
or –
2
,–
4
or –
3
,–
4
. Hence a =
≠(–
i
+ –
4
) for i =1, 2, 3 and the corresponding zero a
2
of r(f) is one of the —
i
.
Notice that the relation –
1
+ –
2
+ –
3
+ –
4
=0shows that the particular role of –
4
in the notations is not essential – the numbers —
i
do not change if we replace –
4
by any other solution of f(x)=0.
Lemma 2.2. The discriminants of the polynomials f and r(f ) are equal. In par-
ticular, if a quartic polynomial f(x) is separable, then its resolvent r(f)(x) is also
separable.
Proof. With the notations as in Lemma 2.1, we have
(r(f)) = (—
1
≠ —
2
)
2
(—
2
≠ —
3
)
2
(—
3
≠ —
1
)
2
=
(–
1
≠ –
2
)
2
(–
1
≠ –
4
)
2
(–
2
≠ –
3
)
2
(–
2
≠ –
4
)
2
(–
3
≠ –
1
)
2
(–
3
≠ –
4
)
2
=(f).
Thus (r(f)) ”=0if and only if (f) ”=0, that is, all –
i
are different if and only
if all —
i
have this property.
2.4 How to decide that a cubic or a quartic has a linear factor?
A cubic polynomial f(x)=x
3
+ px + q is reducible over a field K if and only if
it has a linear factor over K. Thus reducibility means that the polynomial has a
zero in K. Observe that this is not the case for polynomials of higher degree than
3. For example, f(x)=x
4
+4does not have real zeros, but x
4
+4=(x
2
≠ 2x +
2)(x
2
+2x + 2), that is, the polynomial is reducible over R as well over Q.
If we have a polynomial with coefficients in an arbitrary field K, we do not have a
general method by which we could decide whether or not the equation f(x)=0has
a solution in K. The situation is somewhat easier when we have a polynomial with
rational coefficients (which already gives a possibility to exemplify many results
concerning Galois groups in a non-trivial way). Usually one uses an elementary
result saying that if f(x) is a polynomial with integer coefficients and a rational
number – = a/b,wherea, b are relatively prime integers, is a solution of the
equation f(x)=0,thena divides f(0) and b divides the highest coefficient of
f(x). We om it an easy proof. Thus rational solutions must be integer if the highest
coefficient is 1 and all divide f(0) (the “variable-free" term of f(x)). It is still a
serious numerical problem to find all the divisors of the integer f(0) (if it is big),
but there is a lot of good “pedagogical" examples of Galois groups when finding
the divisors of f(0) does not create any problem.
2.5 How to decide that a quartic is reducible?
Proposition 2.1 answers this question as regards factorization into a product of two
quadratic polynomials (the possibility of factorization as a product of a linear and
a cubic polynomial is discussed above in 2.4). For convenience, we recall:
106 Juliusz BrzeziÒski Normat 3-4/2011
Proposition 2.1. Let f(x)=x
4
+px
2
+qx+ r be a polynomial with coefficients in
a field K without zeros in K. Then f is reducible in K (a product of two quadratic
polynomials with coefficients in K) if and only if
(a) q ”=0and the resolvent r (f ) has a zero, which is a square in K,
or
(b) q =0and the resolvent r(f) has two zeros, which are squares in K or ” = p
2
≠4r
is a square in K.
Proof. If f has no zeros in K and is reducible, then it must be a product of two
quadratic polynomials as in (4). The computations leading from (4) to (6) clearly
show that t = a
2
is a zero of the resolvent r(f ), that is, the resolvent has a zero
which is a square in K.
If q ”=0and the resolvent has a zero t = a
2
, a œ K,thena ”=0. Using the system
(5), we find b, b
Õ
œ K and, conse quently, a factorization of f(x) as a product of two
quadratic polynomials. This completes our proof of the first case (a).
In the case (b), if there is a factorization with a =0(of course, a square in K –the
first zero of the resolvent, which is a square in K), then the quadratic polynomial
t
2
+ pt + r has zeros in K, namely, ≠b, ≠b
Õ
, so ” = p
2
≠4r must be a square in K.
Conversely, if ” is a square in K, then this quadratic polynomial has some zeros
≠b, ≠b
Õ
and we get a factorization of f(x) with a =0.
If in the case (b), we have a factorization with a ”=0,thent = a
2
is a second zero of
the re solvent, which is a square in K. Conversely, assume that the resolvent r(f) has
two zeros which are squares in K (these two may be equal to 0). We find easily that
the zeros of r(f)(t)=t
3
+2pt
2
+(p
2
≠4r)t are: —
1
=0,—
2
= ≠p+2
Ô
r, —
3
= ≠p≠2
Ô
r
(observe that
Ô
r œ K). Of course, 0 is one of the squares. If the second is —
2
,we
easily check that we get a factorization of f(x) in K[x] noting that:
f(x)=x
4
+ px
2
+ r =(x
2
+
Ô
r)
2
≠ —
2
x
2
and similarly for —
3
,when—
3
is a square in K.
We note as a corollary an observation made in the last part of the proof:
Corollary 2.1. If f (x)=x
4
+ px
2
+ r, then 0 is always one of the zeros of the
resolvent r(f ) and all three zeros of r (f ) are in K if and only if
Ô
r œ K. Moreover,
the discriminant of f(x)(and r(f)(t)) is
(f) = 16r(p
2
≠ 4r)
2
,
so K(
Ô
) = K(
Ô
r).
Proof. Use the general formulae on (f) from Theorem 1.3 in the case, when
q =0.
Normat 3-4/2011 Juliusz BrzeziÒski 107
2.6 How to describe the splitting fie ld of a quartic?
The following description of a relation between the splitting field of a quartic poly-
nomial and the splitting field of its resolvent is the core of the presentation in this
text. Denote, as before, the solutions of the equation f(x)=x
4
+px
2
+qx +r =0by
–
1
,–
2
,–
3
,–
4
, and the solutions of the resolvent r(f )(t)=t
3
+2pt
2
+(p
2
≠4r)t≠q
2
=
0 given by Lemma 2.1, by —
1
,—
2
,—
3
(in some field K containing K). The n the split-
ting fields of these two polynomials are related in the following way:
Theorem 2.1. The splitting field K
f
over K of a quartic separable polynomial
f(x) œ K[x] can be obtained from the splitting field K
r ( f )
of its resolvent r(f) over
K by adjunction of one arbitrary solution of the equation f(x)=0,thatis,
K
f
= K(–
1
,–
2
,–
3
,–
4
)=K
r ( f )
(–
i
),
where i =1, 2, 3, 4.
Proof. It is clear that
K
r ( f )
(–
i
)=K(—
1
,—
2
,—
3
,–
i
) ™ K(–
1
,–
2
,–
3
,–
4
)=K
f
.
In order to prove the converse inclusion, choose i =4and consider the equations:
f
i
(x)=(x + –
4
)
2
≠ —
i
for i =1, 2, 3. This quadratic polynomial has its coefficients in the field K
r ( f )
(–
4
).
Assume that f
i
(x) is irreducible in this field. By Lemma 2.1, it has a zero x = –
i
,
which is a zero of the polynomial f(x). Hence f
i
(x) divides f(x) (as an irreducible
polynomial having a common zero). Thus one more of the numbers –
1
,–
2
,–
3
,–
4
different from –
i
is a zero of f
i
(x),say,–
j
,wherej ”= i.Then
f
i
(–
i
)=f
i
(–
j
)=0,
that is,
(–
i
+ –
4
)
2
=(–
j
+ –
4
)
2
for j ”= i, which says that —
i
= —
j
– a contradiction according to Lemma 2.2.
Thus the polynomial is reducible in the field K
r ( f )
(–
4
) and its zero –
i
belongs
it. This holds for i =1, 2, 3 (and, of course, also for i =4) and shows that also
K
f
™ K
r ( f )
(–
4
). Together with the converse inclusion, we get K
f
= K
r ( f )
(–
4
).
108 Juliusz BrzeziÒski Normat 3-4/2011
2.7 Proof of Theorem 1.1: Galois groups of irreducible cubics
Proof. Since f(x) is irreducible, the Galois group is a transitive subgroup of the
symmetric group S
3
, that is, it is either A
3
or S
3
(see Appendix). Hence, the split-
ting field K
f
has degree either 3 or 6 over K. The square root of the discriminant
Ô
is a non-zero element of K
f
, since the polynomial f(x) has different zeros.
If
Ô
œ K, then the Galois group must consist of even permutations, since odd
permutations change the sign of
Ô
. Hence, G(K
f
/K)=A
3
and K
f
= K(–
i
) for
any zero –
i
, i =1, 2, 3, of f(x).
If
Ô
”œ K, then the Galois group must consist of both even and odd permutations
(see Proposition 2.1). Hence the Galois group must be S
3
and the degree of K
f
over K is 6.
Notice that in the second case ab ove, the extension K(
Ô
) has degree 2 over K
and K
f
= K(
Ô
,–
i
) for any zero –
i
, i =1, 2, 3, of f(x). In fact, the polynomial
f(x) is still irreducible over K(
Ô
) (as the degree of this field over K is 2 and
the degree of the irreducible over K polynomial f(x) is 3 – see 2.10), so the degree
of K
f
= K(
Ô
,–
i
) over K(
Ô
) is 3. Hence, we have the following analogue of
Theorem 2.1:
Corollary 2.2. If f(x) œ K[x] is an irreducible polynomial of degree 3, then K
f
=
K(
Ô
,–
i
), where –
i
, i =1, 2, 3 is any zero of f(x).
2.8 Proof of Theorem 1.3: Galois groups of irreducible quartics
Proof. We consider 3 cases:
Case 1: The resolvent r(f) has not zeros in K.
This means that r(f ) is an irreducible cubic over K. According to Theorem 1.1,
the splitting field K
r ( f )
has degree 3 or 6 over K depending on the discriminant
=(r(f)) = (f): the first c ase when
Ô
œ K, and the second, when
Ô
”œ K.
If
Ô
œ K (so the degree of K
r ( f )
over K is 3), then the polynomial f (x) of degree 4
is s till irreducible over K
r ( f )
(see 2.10). Hence the splitting field K
f
= K(
Ô
,–
1
),
where –
1
is any zero of f(x) (see 2.6), has degree 4 over K
r ( f )
. Thus the degree of
K
f
over K equals 12, which means that the Galois group G(K
f
/K) is a subgroup
of order 12 of S
4
. There is only one such subgroup (see the Appendix) and it is the
alternating group A
4
of all even permutations of 1, 2, 3, 4.
If
Ô
”œ K, then the Galois group G(K
f
/K) must contain both even and odd
permutations according to Proposition 2.1 and its order is divisible by 6, since the
splitting field K
f
over K contains a subfield K
r ( f )
of degree 6 over K. Moreover,
the Galois group is transitive, since f(x) is irreducible over K. There is only one
such subgroup of S
4
and it is S
4
itself (see the Appendix concerning the claim that
S
4
is the only transitive subgroup of S
4
of order divisible by 6).
Case 2: The resolvent r(f) has exactly one zero — = —
1
=(–
1
+ –
4
)
2
in K.
Normat 3-4/2011 Juliusz BrzeziÒski 109
First of all, let us notice that in this case, we have
Ô
”œ K. In fact, the splitting
field of the resolvent K
r ( f )
= K(
Ô
,—)=K(
Ô
) must be bigger than K,since
otherwise all the zeros of the resolvent are in K.Thus
Ô
can not belong K,sothe
Galois group G(K
f
/K) must contain both even and odd permutations according
to Proposition 2.1.
Which permutations of the zeros –
1
,–
2
,–
3
,–
4
of f(x)=0may belong to G(K
f
/K)?
Since — = —
1
=(–
1
+ –
4
)
2
is in K, the permutations in G(K
f
/K) must fix this
element. We see immediately that such permutations are all in the group:
D
4
= V
4
fi{(1, 2, 4, 3), (1, 3, 4, 2), (1, 4), (2, 3)},
where
V
4
= {(1), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}
(remember that –
1
+ –
4
= ≠(–
2
+ –
3
) and see the Appendix concerning the nota-
tions).
All other permutations in S
4
map — = —
1
on different elements s ince (1, 2) maps
—
1
=(–
1
+ –
4
)
2
onto —
2
and (1, 3) onto —
3
. Thus the 16 products in the cosets
(1, 2)D
4
and (1, 3)D
4
are different and map —
1
onto either —
2
or —
3
.
Assume now that —
1
”=0,soK(
Ô
—)=K(–
1
+ –
4
) is a quadratic extensions of K.
Since the polynomial f(x) is irreducible, the Galois group G(K
f
/K) must contain
at least 4 elements. As a subgroup of D
4
it must consist of 4 or 8 permutations in
D
4
. The group D
4
is the square group and contains 3 subgroups of order 4 (see
the Appendix on the group S
4
and its s ubgroups). Assume that G(K
f
/K) has 4
elements. It can not be the subgroup V
4
, since it only has even permutations all
fixing
Ô
, which however is not in K. It can not be
V
Õ
4
= {(1), (1, 4), (2, 3), (1, 4)(2, 3)},
(see the Appendix for the notations), since this group has –
1
+ –
4
as fixed element
and this is also an element not in K.ThusG(K
f
/K) must be the third group:
C
4
= {(1), (1, 2, 4, 3), (1, 3, 4, 2), (1, 4)(2, 3)},
which is cyclic. But this group only has one subgroup of order 2, so the quadratic
fields K(
Ô
) and K(
Ô
—) (over K) must be equal, that is,
— œ K.Ifthese
two fields are different, that is,
— ”œ K, it remains the only possibility that the
Galois group has order 8 and consists of all permutations in D
4
.
It may happen that — =0, that is, –
1
+ –
4
=0(and, consequently, –
2
+ –
3
=0).
As we noted before (see Remark 1), it happens if and only if the polynomial f(x) is
biquadratic (that is , q =0). Then we have to find another field extension instead of
110 Juliusz BrzeziÒski Normat 3-4/2011
K(
Ô
—),sincenowK(
Ô
—)=K. We take instead K(
Ô
”),sincef(x) is irreducible,
so
Ô
” ”œ K. Notice that
” = p
2
≠ 4r =(–
2
1
≠ –
2
2
)
2
(take into account that –
4
= ≠–
1
and –
3
= ≠–
2
,sop = ≠(–
2
1
+–
2
2
) and r = –
2
1
–
2
2
).
The arguments are similar as in the previous case. As before, the first two groups
with 4 elements are impossible, since the first fixes
Ô
and the other one,
Ô
” =
–
2
1
≠ –
2
2
(this should be checked case by case taking into account the equalities
–
4
= ≠–
1
and –
3
= ≠–
2
). Thus, if the Galois group G(K
f
/K) has order 4, it
must be cyclic. But in this case, the fields K(
Ô
) and K(
Ô
”) must be equal, so
Ô
” œ K. Otherwise, if they are not equal, that is,
Ô
” ”œ K, then the Galois
group has order 8 and contains all permutations in the square group (in this case,
D
4
above).
Case 3: The resolve nt has all three zeros in K.SincenowK
r ( f )
= K,thesplitting
field K
f
= K(–),where– is any zero of f(x) according to Theorem 2.1. Since f(x)
is irreducible, this field has degree 4 over K. Since all —
i
=(–
i
+ –
4
)
2
are in K,
they are fixed by the four permutations in V
4
, which is the Galois group G(K
f
/K)
in this case.
2.9 How to find Galois groups of reducible quartics?
Once we know that the polynomial f(x) is reducible (but without zeros in K), we
can easily describe its Galois group:
Proposition 2.2. Let f(x)=x
4
+ px
2
+ qx+r be reducible in K but without zeros
in K. Then
G(K
F
/K)=
;
V
4
when r(f) has only one zero in K
C
2
when r(f) has all its zeros in K
Proof. Assume that
f(x)=x
4
+ px
2
+ qx + r =(x
2
≠ ax + b)(x
2
+ ax + b
Õ
),
where a, b, b
Õ
œ K, ” = a
2
≠ 4b and ”
Õ
= a
2
≠ 4b
Õ
are not squares in K (since
otherwise, the polynomial f(x) has zeros in K). The zeros of f(x) are a/2 ±
Ô
”
and ≠a/2 ±
Ô
”
Õ
. The splitting field of f(x) over K is K
f
= K(
Ô
”,
Ô
”
Õ
).
The zeros of the resolvent r(f) are according to Lemma 2.1: a
2
, (
Ô
” ±
Ô
”
Õ
)
2
=
” + ”
Õ
± 2
Ô
””
Õ
.
The resolvent r(f) has only one zero in K if and only if
Ô
””
Õ
”œ K,whichis
equivalent to K(
Ô
”) ”= K(
Ô
”
Õ
), that is, K
f
= K(
Ô
”,
Ô
”
Õ
) has degree 4 over K.
The Galois group G(K
f
/K), in this case, is of course isomorphic to V
4
–theKleins
four-group, since all the automorphisms are given by
Ô
” ‘æ ±
Ô
” and
Ô
”
Õ
‘æ ±
Ô
”
Õ
.
Normat 3-4/2011 Juliusz BrzeziÒski 111
(Notice however that as permutation group, it is not V
4
, which in our notations,
denotes a transitive copy of Klein’s four group in S
4
).
The resolvent r(f) has 3 zeros in K if and only if
Ô
””
Õ
œ K, which is equivalent to
K(
Ô
”)=K(
Ô
”
Õ
), that is, K
f
= K(
Ô
”). Thus in this case the splitting field K
f
is
quadratic over K and its Galois group is C
2
– cyclic of order 2.
In fact, when we tes t irreducibility using Proposition 2.1, we can factor f (x) into a
product of two quadratic polynomials (see 2.5 for details). If we already have such
a factorization, then it is useful to note the following direct corollary from the last
proof:
Proposition 2.3. Let
f(x)=x
4
+ px
2
+ qx + r =(x
2
+ ax + b)(x
2
≠ ax + b
Õ
)
be a factorization of f over K, ” = a
2
≠ 4b and ”
Õ
= a
2
≠ 4b
Õ
. Then
G(K
f
/K)=
;
V
4
when
Ô
””
Õ
”œ K
C
2
when
Ô
””
Õ
œ K
2.10 A useful Lemma
Several times in this text, we use the following easy result:
Lemma 2.3. Let L be a finite field extension of a field K and let f(x) œ K[x] be
an irreducible polynomial. If the degrees of f(x) and L over K are relatively prime,
then the polynomial f(x) is still irreducible over L.
Proof. Let n =deg(f ) and r =[L : K] (the degree of the field extension). Let
– be a zero of f(x) in a field containing L.Then[K(–):K]=n (since f(x) is
irreducible over K) and let [L(–):L]=m. Of course, we have m Æ n,sincef(x)
has degree n and may be reducible over L.ThefieldL(–) contains the field K(–)
as well as the field L.
L(–)
m
K(–)
n
L
r
K
Since n and r are relatively prime and both divide [L(–):K],theproductnr
also divides [L(–):K]=[L(–):L][L : K]=mr.Thusn divides m, and since
m Æ n,wehavem = n . This tells us that f(x) is also irreducible over L,since
[L(–):L]=n.
112 Juliusz BrzeziÒski Normat 3-4/2011
2.11 How to solve cubic equations?
We want to solve the equation
x
3
+ px + q =0.(8)
Compare the last equality with the well-known identity:
(a + b)
3
≠ 3ab(a + b) ≠(a
3
+ b
3
)=0(9)
and choose a, b in such a way that:
p = ≠3ab,
q = ≠(a
3
+ b
3
).(10)
If a, b are so chosen, then x = a + b is a solution of the equation (8). The system
a
3
+ b
3
= ≠q,
a
3
b
3
= ≠
p
3
27
,
shows that a
3
,b
3
are solutions of the quadratic equation:
t
2
+ qt ≠
p
3
27
=0.
Solving this equation gives two solutions: t
1
= a
3
and t
2
= b
3
. Then we choose a, b,
which satisfy (10) and find x = a + b. An (impressing) formula, which hardly needs
to be memorized (it is much easier to memorize the presented method based on
the well-known binomial identity (9)) is thus the following:
x
1
= a + b =
3
Û
≠
q
2
+
Ú
q
2
4
+
p
3
27
+
3
Û
≠
q
2
≠
Ú
q
2
4
+
p
3
27
Notice that a, b are simply third roots of t
1
,t
2
, but we have to be a little cautious
since both a and b my be chosen in 3 different ways.
We leave a discussion of different choices of a, b as an easy exercise, but notice
that in order to solve a cubic equation, it is sufficient to find only one root x
1
and
solve a quadratic equation after dividing the cubic by x ≠x
1
. The above expression
(in different notations) was first published by Gerolamo Cardano in his book “Ars
Magna” in 1545 and is known today as Cardano’s formula.
Normat 3-4/2011 Juliusz BrzeziÒski 113
2.12 How to solve quartic equations?
Essentially, we already know the answer – any quartic polynomial with coefficients
in a field K can be split into a product of two quadratic polynomials according to
(4) (with a
Õ
= ≠a):
(11) f(x)=x
4
+ px
2
+ qx + r =(x
2
+ ax + b)(x
2
≠ ax + b
Õ
).
where a, b, b
Õ
are in a field containing K. We know that if we take a as square root
of any nonzero solution of the resolvent equation
r(f)(t)=t
3
+2pt
2
+(p
2
≠ 4r)t ≠ q
2
=0,
the system (5), gives us b and b
Õ
. Thus we factorize the quartic and solve two
quadratic equations in order to find all zeros of f(x).Thisisaverysimplemethod
and very easy to remember if we only know how to deal with a cubic equation (see
2.11). Observe that the res olvent equation always has a nonzero solution, since 0
is the only (triple) solution if and only if p = q = r =0, that is, f(x)=x
4
.
This is in principle the original method given by L. Ferrari
2
in the middle of 16th
century and published in Cardano’s book mentioned above. But the formal lan-
guage of algebra was of course different during Cardano’s and Ferrari’s time, so the
solution method was presented rather as recipe how to transform the equation in
order to get a solution. Today, it is usually presented in the following way. One can
start with the general equation
x
4
+ mx
3
+ px
2
+ qx + r =0,(12)
– it is even not important to assume that m =0(but it could be achieved substi-
tuting x = y ≠
m
4
). It is easy to check that the above equation can be rewritten in
the following way:
3
x
2
+
m
2
x +
⁄
2
4
2
≠
53
m
2
4
+ ⁄ ≠ p
4
x
2
+
3
m⁄
2
≠ q
4
x +
3
⁄
2
4
≠ r
46
=0(13)
We simply want to find a number ⁄ such that the quadratic polynomial in square
brackets is a square of a first degree polynomial. This is achieved when the dis-
criminant of the quadratic polynomial in the square bracket equals 0, that is,
3
m⁄
2
≠ q
4
2
≠ 4
3
m
2
4
+ ⁄ ≠ p
43
⁄
2
4
≠ r
4
=0(14)
2
Lodovico Ferrari (1522-1565) was an Italian mathematician who was a servant, later a student
and finally a successor at the Universisty of Pavia of G. Cardano.
114 Juliusz BrzeziÒski Normat 3-4/2011
This is a cubic equation with respect to ⁄. Solving it (e.g. according to 2.11), we
get a root ⁄, which gives a poss ibility to split the quartic polynomial (12) into a
product of two quadratic polynomials and then solve two quadratic equations.
3 APPENDIX: The group S
4
and their subgroups
As a permutation group, the Galois group G(K
f
/K) of an irreducible polynomial
f(x) œ K[x] has an important property of transitivity. A permutation group G ™
S
n
is called transitive if for any pair i, j œ{1, 2,...,n} there is a permutation
‡ œ G such that ‡(i)=j. Of course, this property is valid for Galois groups of
irreducible polynomials, since according to well-known properties of automorphisms
of splitting fields, there is always an automorphism which maps any given ze ro of
f(x) onto any other zero of this polynomial.
The symmetric group S
3
of all permutations of 1, 2, 3 consists of 3! = 6 elements. We
can represent each permutation as an isometry of the plane mapping the equatorial
triangle with vertices in the points 1, 2, 3 on itself. If {a, b, c} = {1, 2, 3},then
there are 3 rotations: the identity (1), the rotations ±120
0
: (1, 2, 3), (1, 3, 2) and 3
symmetries in the three heights of the triangle: (1, 2), (2, 3), (1, 3). There are only
two transitive subgroups of S
3
: the group S
3
itself and the subgroup of the rotations
(all even permutations) A
3
= {(1), (1, 2, 3), (1, 3, 2)}. All these facts are very easy
to check (e.g. by listing all the subgroups of S
3
).
The symmetric group S
4
of all permutations of 1, 2, 3, 4 consists of 4! = 24 elements.
We can represent each permutation as an isometry of the space mapping the tetra-
hedron with vertices in the points 1, 2, 3, 4 on itself. If {a, b, c, d } = {1, 2, 3, 4},
then each non-identity permutation can be written as a cycle or a composition of
them:
6symmetries(a, b) of order 2: in the planes through the vertices c, d and the middle
of the side between a and b;
8 rotations (a, b, c) of order 3: around the axis through the vertex d perpendicularly
to the plane through the points a, b, c;
3 rotations (a, b)(c, d) of order 2: 180 degrees around the axis through the middles
of the sides a, b and c, d. Notice that these rotations together with the identity (1)
form a transitive group of order 4. It is a transitive presentation of Klein’s four
group, which is usually denoted by V
4
, that is,
V
4
= {(1), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}.
6cycles(a, b, c, d) of order 4: compos itions of the rotation (a, b, c) with the symme-
try (c, d).
Together with the identity (1), we have 24 all possible isometric mappings of the
tetrahedron on itself. Notice that the even permutations are exactly the 12 rota-
tions, and the odd permutations are the 6 symmetries and the 6 compositions of a
rotation of order 3 and a symmetry.
Normat 3-4/2011 Juliusz BrzeziÒski 115
In order to describe all transitive subgroups of S
4
, notice that in any subgroup
G, which contains at least one odd permutation ‡, the half of the elements are
odd permutations, and the other half are even. In fact, if G
0
denotes all even
permutations in G,thenG
0
is a (normal) subgroup of G and G = G
0
fi‡G
0
,since
every permutation · in G is either even or, if it is odd, then ‡
≠1
· is even (that is,
in G
0
).
Using this observation, we first note that the group S
4
has exactly one (normal)
subgroup of order 12. This is the subgroup A
4
consisting of even p ermutations
(rotations of the te trahedron)
3
. In fact, if G is a subgroup of S
4
of order 12 and
contains at least one odd permutation, then the intersection A
4
flG is a subgroup
of A
4
of order 6. Such a subgroup of order 6 is not cyclic (there are no elements
of order 6 in S
4
), so it must be isomorphic with S
3
. Hence, it contains 3 elements
of order 2. The only even permutations of order 2 in S
4
are the 3 rotations 180
0
,
which together with the identity form the group V
4
of order 4. This is of course a
contradiction, since a group of order 4 can not be a subgroup of a group of order
6. Thus S
4
has only one subgroup of order 12 – the one consisting of all even
permutations.
Next, the group S
4
may have subgroups of order 8. There are in fact 3 subgroups
of order 8. All are isomorphic with the square group D
4
and are transitive. In order
to prove this, assume that G is a subgroup of order 8. Then it must consist of both
even and odd permutations – all can not be even, since a group of order 8 can not
be a subgroup of a group of order 12. The subgroup G
0
of G consisting of the even
permutations (that is, rotations of the tetrahedron) must be G
0
= V
4
,sincethe
remaining rotations all have order 3 (can not belong to a group of order 4). The
odd permutations in G can not all be of order 2. Those are exactly the symmetries
of the tetrahedron. Between 4 such symmetries, there are at least 2 which shift the
same vertex (of four possible), that is, they have form (a, b) and (a, c).Thenthe
group G contains (a, b)(a, c)=(a, c, b), which as an element of order 3. This order
is not allowed by G.ThusG must contain an odd permutation ‡ of order 4. An
easy direct computation shows that there are exactly 3 possibilities for ‡V
4
,which
give 3 possibilities for G:
D
4
= V
4
fi{(1, 2, 4, 3), (1, 3, 4, 2), (1, 4), (2, 3)},
D
Õ
4
= V
4
fi{(1, 2, 3, 4), (1, 4, 3, 2), (1, 3), (2, 4)},
D
ÕÕ
4
= V
4
fi{(1, 3, 2, 4), (1, 4, 2, 3), (1, 2), (3, 4)}.
It is clear that these groups are transitive. Each group gives a description of all
isometries of a square corresponding to a numbering of its vertices a, b, c, d according
to the rotations of square given by the elements of order 4 belonging to it.
3
In general, the subgroup consisting o f all even permutations in S
n
is denoted by A
n
and
called the alternated group of degree n.Itsorderisn!/2.
116 Juliusz BrzeziÒski Normat 3-4/2011
There are no transitive subgroups G of S
4
of order 6. In fact, such a subgroup
can not be cyclic, since there are no e leme nts of order 6 in S
4
.Thusitmust
be isomorphic to S
3
. As we know such a group has 3 elements of order 2 and 2
elements of order 3. Between the elements of order 2 must be at least 2 symmetries
(otherwise G
0
is a subgroup of G, which is impossible). They must shift a common
vertex a. Otherwise, they are of the form (a, b), (c, d) with different a, b, c, d.Then
(1), (a, b), (c, d), (a, b)(c, d) is a subgroup of G, which is impossible. If (a, b) and
(a, c) are in G, then it is easy to check that G is the group of all permutations of
a, b, c. It is not transitive on the set {1, 2, 3, 4}.
Finally, there is only one transitive subgroup of order 4 – the group V
4
. In fact, it
is easy to check that the subgroups generated by the elements of order 4 are not
transitive. A non-cyclic subgroup of order 4 must contain 3 elements of order 2. A
similar argument to that given above in connection with the subgroups of order 6
shows that it is impossible to get a transitive group with two symmetries. Thus we
can only have the rotations giving V
4
.
Of course, the subgroups of order 2 are never transitive.
Summarizing, we have the following list of transitive subgroups of S
4
:
S
4
,A
4
,D
4
,D
Õ
4
,D
ÕÕ
4
,V
4
.
The group V
4
is a subgroup of all these groups. Notice that the groups D
4
,D
Õ
4
,D
ÕÕ
4
are all isomorphic. They consist of all isometries of a square corresponding to a
numbering of its vertices a, b, c, d (according to the rotations given by the elements
of order 4 belonging to it). Recall that such a group contains 3 subgroups of order
4: V
4
(the rectangle group), the cyclic group of the rotations of the square:
C
4
= {(1), (a, b, c, d), (a, d, c, b), (a, c)(b, d)}
and a non-transitive representation of Klein’s four group (the romb group):
V
Õ
4
= {(1), (a, b), (c, d), (a, b)(c, d)}.
REFERENCES
[B] J. BrzeziÒski, Galois groups of quartic polynomials (in Polish), Wiadomoúci
Matematyczne X(1968), 133–143.
[C] D. Cox, Galois Theory, Wiley-Interscience, 2012.
[L] S. Lang, Algebra, Addison–Wesley, 1993.
[N] T. Nagell, Lärobok i algebra, Hugo Gebers Förlag, Uppsala 1949.
[T] N.G. Tschebotaröw, H. Schwerdtfeger, Grundzüge Der Galois’schen Theorie,
P. Nordhoff N.V., Groningen-Djakarta, 1950.
